One move, then another
A combined transformation is two transformations done one after the other: the second acts on the image made by the first. If the object is A, the first image is A', and the second is A″, read "A double prime".
Take the triangle A with corners (1, 1), (3, 1) and (3, 2). First reflect it in the y-axis, which changes the sign of every x-coordinate: A' has corners (−1, 1), (−3, 1) and (−3, 2). Then reflect A', not A, in the x-axis, which changes the sign of every y-coordinate: A″ has corners (−1, −1), (−3, −1) and (−3, −2).
The first move: A reflected in the y-axis, the gold mirror line, gives A'.
The second move: A' reflected in the x-axis, now the mirror line, gives A″.
One single move does the same
Follow a general point through both moves: . Both signs have changed, and that is the rule for a half turn about the origin, from Rotation. So reflecting in the y-axis and then in the x-axis is the same as one rotation of 180° about the origin.
Check with a corner: (3, 2) goes to (−3, 2) and then to (−3, −2), and a half turn sends (3, 2) straight to (−3, −2).
The single move is a rotation, not a reflection. One reflection turns a shape over, so its corners go round the other way; a second reflection turns it back again.
A half turn about the origin sends A straight to A″: (1, 1), (3, 1) and (3, 2) go to (−1, −1), (−3, −1) and (−3, −2).
Swapping this pair changes nothing
Do the same two reflections the other way round: the x-axis first, then the y-axis. . The result is the same half turn. Each of these two reflections changes one sign, a different one, and it does not matter which sign is changed first.
In the other order, the first move reflects A in the x-axis. Reflecting this image in the y-axis then lands on the same A″ as before.
For many pairs, the order matters
Now combine a reflection in the y-axis with a rotation of 90° counterclockwise about the origin, which sends (x, y) to (−y, x).
Reflect first, then rotate: . Rotate first, then reflect: . The two results are different. The corner (3, 1) ends at (−1, −3) in the first order and at (1, 3) in the second.
Each result is a single reflection: is the reflection in y = −x, and is the reflection in y = x. So the same two moves, done in the two orders, give reflections in two different mirror lines.
Where each corner of A ends up. Reflecting in the y-axis and then rotating 90° counterclockwise gives the first column; rotating first and then reflecting gives the second.
Rotating A 90° counterclockwise about the origin and then reflecting the result in the y-axis lands on the reflection of A in y = x: (1, 1), (1, 3) and (2, 3). The corner (1, 1) is on the mirror line.
Two reflections in parallel mirrors
Reflect in the line y = 1, and then in the line y = 3. The mirrors are parallel, 2 apart. The point (1, 0) is 1 below y = 1, so its first image is 1 above it, at (1, 2). That point is 1 below y = 3, so its second image is 1 above it, at (1, 4). It has moved 4 straight up.
Follow a general point. A point at height y is y − 1 above the first mirror (a negative distance above means below), so its image is y − 1 below it, at height 1 − (y − 1) = 2 − y. That is 3 − (2 − y) = 1 + y below the second mirror, so the second image is 1 + y above it, at height 3 + 1 + y = y + 4. The x-coordinate never changes. So every point goes from (x, y) to (x, y + 4): the two reflections make one translation by [0, 4].
The distance moved is 4, twice the gap of 2 between the mirrors, and the direction is at right angles to the mirrors, from the first mirror toward the second. With the order swapped, the translation is [0, −4].
The gold line is the first mirror, y = 1, and the white line the second, y = 3. The point (1, 0) goes to (1, 2) and then to (1, 4). The point (4, 2.5), between the mirrors, goes to (4, −0.5) and then to (4, 6.5). Both end 4 higher than they began.
Two reflections in mirrors that cross
When the two mirror lines cross, the two reflections make a rotation about the crossing point, through twice the angle between the mirrors.
Here is why. Each reflection keeps every point the same distance from the crossing point, so the combination only turns points about it. Measure angles from the first mirror, counterclockwise, and let the second mirror be at angle . A point at angle is reflected in the first mirror to angle . That is short of the second mirror, so its reflection in the second mirror is beyond it, at angle . Every point has turned through .
The y-axis and the x-axis cross at 90°, so reflecting in one and then the other is a rotation of 2 × 90° = 180°, the half turn that the two axis reflections gave. Mirrors at 45° make a quarter turn, 90°, and mirrors at 60° make a turn of 120°. The direction of the turn is from the first mirror toward the second, so swapping the order turns the other way.
The usual mistakes
Applying the second move to the object instead of the first image. The second move acts on A', so for the two axis mirrors (3, 2) ends at (−3, −2), not at (3, −2).
Describing two reflections as a reflection. The second reflection turns the shape back over, so the single move is a rotation or a translation.
Calling the half turn a quarter turn. The two axis reflections only change signs; a quarter turn also swaps the coordinates.
Assuming the order never matters, or always matters. For the two axis reflections it does not; for a reflection and a quarter turn it does. Work out both orders to be sure.
Giving the gap between parallel mirrors as the distance moved. The translation is twice the gap.
Worked example: Two Mirrors of a Kaleidoscope at 45 Degrees, and the Image of an Image
Question In a kaleidoscope, two long mirrors meet along a line at an angle of 45°. Looking down the tube, a grid in millimeters puts the mirrors along the positive x-axis and along the line y = x, meeting at O(0, 0). A triangular bead between them has corners A(4, 1), B(6, 1) and C(5, 2). (a) The bead is reflected in the mirror along the x-axis, and that image is reflected in the line y = x. Find the coordinates of the corners of the final image A''B''C''. (b) Describe fully the single transformation that maps the bead onto A''B''C''. Is the result the same if the bead is reflected in y = x first?
1.Reflecting in the x-axis changes the sign of each y-coordinate: A(4, 1) goes to (4, −1), B(6, 1) goes to (6, −1) and C(5, 2) goes to (5, −2).
Reflecting in the x-axis changes the sign of each y-coordinate. 2.Reflecting in y = x swaps the coordinates: (4, −1) goes to (−1, 4), (6, −1) goes to (−1, 6) and (5, −2) goes to (−2, 5).
Reflecting that image in y = x swaps the two coordinates of each corner. 3.(a) The final image has corners A''(−1, 4), B''(−1, 6) and C''(−2, 5).
(a) The final image has corners A''(−1, 4), B''(−1, 6) and C''(−2, 5). 4.Compare A with A'' from O. OA = √42 + 12 = √17 and OA'' = √(−1)2 + 42 = √17, so they are equally far from O. The gradient of OA is 14 and the gradient of OA'' is 4−1 = −4, and 14 × (−4) = −1, so OA and OA'' are at right angles. The same holds for B and C, and each image is a quarter turn counterclockwise from its corner.
OA = OA'' = √17, and the gradients 14 and −4 multiply to −1: a quarter turn counterclockwise about O. 5.In the other order, y = x first sends A(4, 1) to (1, 4), and then the x-axis sends it to (1, −4). In the same way B ends at (1, −6) and C at (2, −5). Each of these is a quarter turn clockwise about O from its corner.
Reflected in y = x first, the bead ends at (1, −4), (1, −6) and (2, −5): a quarter turn clockwise about O. 6.(b) The single transformation is a rotation of 90° counterclockwise about O(0, 0), which is twice the 45° between the mirrors. Reflecting in y = x first gives a rotation of 90° clockwise about O instead, so the result is not the same: the order of the two reflections matters.
(b) A rotation of 90° counterclockwise about O(0, 0), twice the angle between the mirrors. In the other order it is 90° clockwise, so the order matters.
Answer: (a) A''(−1, 4), B''(−1, 6) and C''(−2, 5); (b) a rotation of 90° counterclockwise about O(0, 0); reflecting in y = x first gives a rotation of 90° clockwise about O, so the order matters
Common mistakes
- Describing the result as a reflection, because it was made from reflections. The corners A, B, C run counterclockwise round the bead, and so do A'', B'', C''; one reflection turns a shape over and reverses that order, and a second reflection turns it back.
- Giving the rotation without its center or direction. A quarter turn clockwise about O puts A at (1, −4), and a quarter turn about another point puts it somewhere else again, so the center, the angle and the direction are all part of the answer.
Worked example: A Mat Between the Parallel Mirrors of a Dance Studio, and Its Image in the Far Mirror
Question A dance studio has mirrors on two opposite walls, which lie along the lines x = 3 and x = 8 on a floor plan marked in meters. A triangular mat on the floor has corners A(4, 2), B(5, 2) and C(4, 4). A dancer sees the mat reflected in the mirror on x = 3, and that image reflected again in the mirror on x = 8. (a) Find the coordinates of the corners A'', B'' and C'' of this second image. (b) Describe fully the single transformation that maps the mat onto the second image, and show that it would be the same wherever the mat lay between the mirrors.
1.In the mirror on x = 3, A(4, 2) is 1 m in front of the line, so its image is 1 m behind it, at (2, 2). In the same way B(5, 2) goes to (1, 2) and C(4, 4) goes to (2, 4). In each case x becomes 6 − x.
In the mirror on x = 3 each corner goes as far behind the line as it was in front: x becomes 6 − x. 2.In the mirror on x = 8, x becomes 16 − x: (2, 2) goes to (14, 2), (1, 2) goes to (15, 2) and (2, 4) goes to (14, 4).
In the mirror on x = 8, x becomes 16 − x, which gives the second image. 3.(a) The second image has corners A''(14, 2), B''(15, 2) and C''(14, 4).
(a) The second image has corners A''(14, 2), B''(15, 2) and C''(14, 4). 4.Each corner has moved 10 m in the x-direction and not at all in the y-direction: A(4, 2) to (14, 2), B(5, 2) to (15, 2) and C(4, 4) to (14, 4). The mat also faces the same way as before, with B'' to the right of A'' and C'' above it, so this is a translation by 100.
Every corner has moved 10 m in the x-direction, and the mat faces the same way: a translation by 100. 5.For a general point (x, y), the first mirror gives (6 − x, y) and the second gives (16 − (6 − x), y) = (x + 10, y). The 10 does not depend on x or on y, and it is twice the 8 − 3 = 5 m between the mirrors.
A general point (x, y) goes to (6 − x, y) and then to (x + 10, y), whatever x and y are. 6.(b) The single transformation is the translation by 100: 10 m at right angles to the mirrors, twice the distance between them. It is the same wherever the mat lies, because every point (x, y) goes to (x + 10, y).
(b) The translation by 100, twice the 5 m between the mirrors, the same for every point between them.
Answer: (a) A''(14, 2), B''(15, 2) and C''(14, 4); (b) the translation by 100, twice the 5 m between the mirrors, the same for every point
Common mistakes
- Expecting the second image to be turned over, because each step was a reflection. The first reflection turns the mat over and the second turns it back, so the second image faces the same way as the mat.
- Giving the translation as 5 m, the distance between the mirrors. The general point moves 10 m: each reflection carries a point to twice its distance from the mirror, and together the two moves add up to twice the gap.